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I can't comment at the moment so if the moderators feel this needs to be converted to a comment, please feel free to do so.

First of all, how do you know that $H^1(X,\mathscr{I}_x(n)) = 0$ for any $n > 0$? Serre vanishing is true for $n >> 0 $ and not for every $n > 0$.

Nor the sake of simplicity I will tackle the case that $x$ is the closed point $V_+(p,x_0)$ and compute $H^0(X,\mathscr{I}_x(n))$ using \v{C}ech cohomology. In our case, the relevant complex looks like

$$ 0 \to (x_0,p)_{x_0}(n) \times (x_0,p)_{x_1}(n) \stackrel{d}{\to} (x_0,p)_{x_0x_1}(n) \to 0$$

where the notation $(x_0,p)_{x_0}(n)$ means the degree $n$ component of the localization $S^{-1}(x_0,p)$ where $S = \{1,x_0,x_0^2,\ldots\}$. I will now identify $\ker d = H^0(X,\mathscr{I}_x(n))$.

Suppose we had a pair $(f/x_0^k, g/x_1^l)$ in the first non-zero term of the complex. WLOG we may assume

1. $f,g$ are respectively not multiples of $x_0,x_1$
2. $\deg f - k = n$ and $\deg g - l = n$.

If our pair mapped to zero under $d$ then $x_1^lf = gx_0^k$ which forces $k = l= 0$.

Conclusion

The kernel of $d$ consists of homogeneous polynomials of degree $n$ in the ideal $(x_0,p)$. This is free of rank $n + 1$ and the good news is it agrees with what you found above! Hopefully from this you can see how to generalize to other closed points.

For the sake of simplicity I will tackle the case that $x$ is the closed point $V_+(p,x_0)$ and compute $H^0(X,\mathscr{I}_x(n))$ using \v{C}ech cohomology. In our case, the relevant complex looks like

$$ 0 \to (x_0,p)_{x_0}(n) \times (x_0,p)_{x_1}(n) \stackrel{d}{\to} (x_0,p)_{x_0x_1}(n) \to 0$$

where the notation $(x_0,p)_{x_0}(n)$ means the degree $n$ component of the localization $S^{-1}(x_0,p)$ where $S = \{1,x_0,x_0^2,\ldots\}$. I will now identify $\ker d = H^0(X,\mathscr{I}_x(n))$.

Suppose we had a pair $(f/x_0^k, g/x_1^l)$ in the first non-zero term of the complex. WLOG we may assume

1. $f,g$ are respectively not multiples of $x_0,x_1$
2. $\deg f - k = n$ and $\deg g - l = n$.

If our pair mapped to zero under $d$ then $x_1^lf = gx_0^k$ which forces $k = l= 0$.

Conclusion

The kernel of $d$ consists of homogeneous polynomials of degree $n$ in the ideal $(x_0,p)$. This is free of rank $n + 1$ and the good news is it agrees with what you found above! Hopefully from this you can see how to generalize to other closed points.

I can't comment at the moment so if the moderators feel this needs to be converted to a comment, please feel free to do so.

First of all, how do you know that $H^1(X,\mathscr{I}_x(n)) = 0$ for any $n > 0$? Serre vanishing is true for $n >> 0 $ and not for every $n > 0$.

Nor the sake of simplicity I will tackle the case that $x = [0:1]$ and compute $H^0(X,\mathscr{I}_x(n))$ using \v{C}ech cohomology. In our case, the relevant complex looks like

$$ 0 \to (x_0)_{x_0}(n) \times (x_0)_{x_1}(n) \stackrel{d}{\to} (x_0)_{x_0x_1}(n) \to 0$$

where the notation $(x_0)_{x_0}(n)$ means the degree $n$ component of the localization $S^{-1}(x_0)$, $S = \{1,x_0,x_0^2,\ldots\}$. I will now identify $\ker d = H^0(X,\mathscr{I}_x(n))$.

Suppose we had a pair $(f/x_0^k, g/x_1^l)$ in the first non-zero term of the complex. WLOG we may assume

1. $f,g$ are respectively not multiples of $x_0,x_1$
2. $\deg f - k = n$ and $\deg g - l = n$.

If our pair mapped to zero under $d$ then $x_1^lf = gx_0^k$ which forces $k = l= 0$.

Conclusion

The kernel of $d$ consists of homogeneous polynomials of degree $n$ in the ideal $(x_0)$. Hopefully from this you can see how to generalize to other closed points.

I can't comment at the moment so if the moderators feel this needs to be converted to a comment, please feel free to do so.

First of all, how do you know that $H^1(X,\mathscr{I}_x(n)) = 0$ for any $n > 0$? Serre vanishing is true for $n >> 0 $ and not for every $n > 0$.

Nor the sake of simplicity I will tackle the case that $x$ is the closed point $V_+(p,x_0)$ and compute $H^0(X,\mathscr{I}_x(n))$ using \v{C}ech cohomology. In our case, the relevant complex looks like

$$ 0 \to (x_0,p)_{x_0}(n) \times (x_0,p)_{x_1}(n) \stackrel{d}{\to} (x_0,p)_{x_0x_1}(n) \to 0$$

where the notation $(x_0,p)_{x_0}(n)$ means the degree $n$ component of the localization $S^{-1}(x_0,p)$ where $S = \{1,x_0,x_0^2,\ldots\}$. I will now identify $\ker d = H^0(X,\mathscr{I}_x(n))$.

Suppose we had a pair $(f/x_0^k, g/x_1^l)$ in the first non-zero term of the complex. WLOG we may assume

1. $f,g$ are respectively not multiples of $x_0,x_1$
2. $\deg f - k = n$ and $\deg g - l = n$.

If our pair mapped to zero under $d$ then $x_1^lf = gx_0^k$ which forces $k = l= 0$.

Conclusion

The kernel of $d$ consists of homogeneous polynomials of degree $n$ in the ideal $(x_0,p)$. This is free of rank $n + 1$ and the good news is it agrees with what you found above! Hopefully from this you can see how to generalize to other closed points.

I can't comment at the moment so if the moderators feel this needs to be converted to a comment, please feel free to do so.

First of all, I have a few comments to make. How do you know that $H^1(X,\mathscr{I}_x(n)) = 0$ for any $n > 0$? Serre vanishing is true for $n >> 0 $ and not for every $n > 0$. For the sake of simplicity I will tackle the case that $x = [0:1]$ and compute $H^0(X,\mathscr{I}_x(n))$ using \v{C}ech cohomology. In our case, the relevant complex looks like

$$ 0 \to (x_0)_{x_0}(n) \times (x_0)_{x_1}(n) \stackrel{d}{\to} (x_0)_{x_0x_1}(n) \to 0$$

where the notation $(x_0)_{x_0}(n)$ means the degree $n$ component of the localization $S^{-1}(x_0)$, $S = \{1,x_0,x_0^2,\ldots\}$. I will now identify $\ker d = H^0(X,\mathscr{I}_x(n))$.

Suppose we had a pair $(f/x_0^k, g/x_1^l)$ in the first non-zero term of the complex. WLOG we may assume

1. $f,g$ are respectively not multiples of $x_0,x_1$
2. $\deg f - k = n$ and $\deg g - l = n$.

If our pair mapped to zero under $d$ then $x_1^lf = gx_0^k$ which forces $k = l= 0$.

Conclusion

The kernel of $d$ consists of homogeneous polynomials of degree $n$ in the ideal $(x_0)$. Hopefully from this you can see how to generalize to other closed points.

I can't comment at the moment so if the moderators feel this needs to be converted to a comment, please feel free to do so.

First of all, how do you know that $H^1(X,\mathscr{I}_x(n)) = 0$ for any $n > 0$? Serre vanishing is true for $n >> 0 $ and not for every $n > 0$.

Nor the sake of simplicity I will tackle the case that $x = [0:1]$ and compute $H^0(X,\mathscr{I}_x(n))$ using \v{C}ech cohomology. In our case, the relevant complex looks like

$$ 0 \to (x_0)_{x_0}(n) \times (x_0)_{x_1}(n) \stackrel{d}{\to} (x_0)_{x_0x_1}(n) \to 0$$

where the notation $(x_0)_{x_0}(n)$ means the degree $n$ component of the localization $S^{-1}(x_0)$, $S = \{1,x_0,x_0^2,\ldots\}$. I will now identify $\ker d = H^0(X,\mathscr{I}_x(n))$.

Suppose we had a pair $(f/x_0^k, g/x_1^l)$ in the first non-zero term of the complex. WLOG we may assume

1. $f,g$ are respectively not multiples of $x_0,x_1$
2. $\deg f - k = n$ and $\deg g - l = n$.

If our pair mapped to zero under $d$ then $x_1^lf = gx_0^k$ which forces $k = l= 0$.

Conclusion

The kernel of $d$ consists of homogeneous polynomials of degree $n$ in the ideal $(x_0)$. Hopefully from this you can see how to generalize to other closed points.